Q 29E - University Physics with Modern Physics

Question

Question: A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0m/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of  $\mathbf{1}\mathbf{.}\mathbf{10}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$? (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is $\mathbf{1}\mathbf{.}\mathbf{62}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Step-by-Step Solution

Answer

a) The coefficient of static friction is, 0.700 and kinetic friction is, 0.471.

b) The push must exerted to give acceleration is, 257.2 N.

c) (i) The magnitude push which cause it to move is, 51.03 N.

   (ii) The acceleration is, $4.95m/s^2$. 

Step 1: Identification of given data

The given data can be listed below as,

• The mass of the crate is, m = 45.0kg.
• The force when crate just begin to move is, $F_s=313 \text{N}$.
• The force to keep crate moving is, $F_k=208 \text{N}$.
• The steady speed of crate is, $v=25.0 \text{m/s}$.

Step 2: Significance of friction

Friction is an opposing force which resist one surface to slide over another surface when two surface are in contact with each other.

Step 3: Determination of coefficient of static friction and kinetic friction.

The expression for the maximum static force can be expressed as,

 $$\begin{gathered} F_s={\mu }_s m g \\ {\mu }_s=\frac{F_s}{m g} \end{gathered}$$

Here ${\mu }_s$ is the coefficient of static friction, $F_s$ is the static force, m is the mass of the crate, and g is acceleration due to gravity.

 

Substitute, 313 N for $F_s$, 45.0 kg for m, $9.8m/s^2$ for g in the above equation.

 $\begin{array}{rcl}{\mu }_s& =& \frac{313 \text{N}}{45.0 \text{kg} \times 9.8 {\text{m/s}}^{\text{2}}}\\ & =& 0.700\end{array}$

Hence, required coefficient of static friction is, 0.700.

 

The expression for the maximum kinetic force can be expressed as,

$$\begin{gathered} F_k={\mu }_k m g \\ {\mu }_k=\frac{F_k}{m g} \end{gathered}$$

Here ${\mu }_k$ is the coefficient of kinetic friction, $F_k$ is the maximum kinetic force.

 

Substitute 208 N for $F_k$, 45.0 kg for m, $9.8m/s^2$ for g in the above equation.

  $\begin{array}{rcl}{\mu }_k& =& \frac{208 \text{N}}{45.0 \text{kg} \times 9.8 {\text{m/s}}^{\text{2}}}\\ & =& 0.471\end{array}$

Hence, required coefficient of kinetic friction is, 0.471.

Step 4: determination of pushing force.

The expression for the pushing force on crate according to Newton’s second law can be expressed as,

$F= m a+{\mu }_k m g$ 

Here m is the mass, a is the acceleration, and ${\mu }_k$ is the coefficient of kinetic friction, and g is the acceleration due to gravity.

 

Substitute 45.0 kg for m, 0.471 for ${\mu }_k$, and $1.10m/s^2$ for a, $9.8m/s^2$ for g in the above equation.

$\begin{array}{rcl}F& =& (45.0 \text{kg}) (1.10 {\text{m/s}}^{\text{2}})+(0.471) (45.0 \text{kg}) (9.8 {\text{m/s}}^{\text{2}})\\ & =& 257.2 \text{N}\end{array}$

Hence, required force is, 257.2 N.

Step 5: Determination of magnetic push and the acceleration on the moon.

Part (i)

The expression for static force on the moon can be expressed as,

 $F_s={\mu }_s m g$

Here ${\mu }_s$ is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity on moon.

Substitute 0.700 for ${\mu }_s$, 45.0 kg for m, and $1.62m/s^2$ for g in the above equation.

 $\begin{array}{rcl}{F}_s& =& (0.700) (45.0 \text{kg}) (1.62 {\text{m/s}}^{\text{2}})\\ & =& 51.03 \text{N}\end{array}$

Hence, required magnitude to move is, 51.03.

 

Part (ii)

The expression for acceleration in this condition can be expressed as,

 $$\begin{gathered} F-{\mu }_k m g=m a \\ a=\frac{F-{\mu }_k m g}{m} \end{gathered}$$

Here F is pushing force, m is the mass, g is the acceleration due to gravity, and ${\mu }_k$ is the coefficient of kinetic friction.

 

Substitute 257.2 N for F, 0.471 for ${\mu }_k$, 45.0 kg for m, $1.62m/s^2$ for g in the above equation.

 data-custom-editor="chemistry" $\begin{array}{rcl}a& =& \frac{257.2 \text{N}-(0.471) (45.0 \text{kg}) (1.62 {\text{m/s}}^{\text{2}})}{45.0 \text{kg}}\\ & =& 4.95 {\text{m/s}}^{\text{2}}\end{array}$

Hence, required acceleration is, $4.95m/s^2$.